Download Electromagnetic Waves from Maxwell`s Equations

MISN-0-210
ELECTROMAGNETIC WAVES FROM MAXWELL’S EQUATIONS
by
Peter Signell
Michigan State University
1. Introduction
a. Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
b. Waves in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1
c. Vector Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
d. Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
ELECTROMAGNETIC WAVES FROM
MAXWELL’S EQUATIONS
2. Vector Derivatives
a. The Gradient Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
b. The Divergence Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
c. The Curl Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
y
`
cB
3. Maxwell’s Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
4. Electromagnetic Waves
a. No Charge, No Current; Waves . . . . . . . . . . . . . . . . . . . . . . . . . . 3
b. Plane-Polarized Monochromatic Waves . . . . . . . . . . . . . . . . . . 4
c. Production by a Radio Transmitter . . . . . . . . . . . . . . . . . . . . . . 5
d. How The Waves Manifest Themselves . . . . . . . . . . . . . . . . . . . 6
x
Acknowledgments. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6
`
E
Glossary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
A. The Curl as a Determinant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
z
B. Proof of a Vector Identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
Project PHYSNET · Physics Bldg. · Michigan State University · East Lansing, MI
1
ID Sheet: MISN-0-210
THIS IS A DEVELOPMENTAL-STAGE PUBLICATION
OF PROJECT PHYSNET
Title: Electromagnetic Waves from Maxwell’s Equations
Author: P. Signell, Michigan State University
Version: 10/18/2001
Evaluation: Stage 0
Length: 1 hr; 20 pages
Input Skills:
1. Vocabulary: charge density (MISN-0-147), current density (MISN0-118), displacement (MISN-0-25), sound waves, wave frequency,
wavelength, waves on strings, wave speed (MISN-0-202), wave
equation (MISN-0-201).
2. Take derivatives of transcendental functions (MISN-0-1).
3. Take scalar and vector products of vectors using Cartesian unit
vectors (MISN-0-2).
Output Skills (Knowledge):
K1. Vocabulary: propagation (of a wave), polarization (direction of),
plane-polarized (wave), monochromatic (wave).
K2. Given Maxwell’s Equations, the “curl-curl” vector identity, and
the definitions of the gradient, divergence, and curl operators, derive the wave equations for electric and magnetic field vectors at
chargeless currentless space-points.
Output Skills (Rule Application):
R1. Given the definitions of the gradient, divergence, and curl operators, verify that a given electromagnetic wave, consisting of coupled electric and magnetic waves, satisfies Maxwell’s Equations.
R2. Given the direction of polarization, direction of propagation, frequency and amplitude of a monochromatic plane-polarized electromagnetic wave, write down the electric and magnetic fields in
vector form. Sketch the situation.
Post-Options:
1. “Energy and Momentum in Electromagnetic Waves,” (MISN-0211).
2. “Brewster’s Law and Polarization,” (MISN-0-225).
The goal of our project is to assist a network of educators and scientists in
transferring physics from one person to another. We support manuscript
processing and distribution, along with communication and information
systems. We also work with employers to identify basic scientific skills
as well as physics topics that are needed in science and technology. A
number of our publications are aimed at assisting users in acquiring such
skills.
Our publications are designed: (i) to be updated quickly in response to
field tests and new scientific developments; (ii) to be used in both classroom and professional settings; (iii) to show the prerequisite dependencies existing among the various chunks of physics knowledge and skill,
as a guide both to mental organization and to use of the materials; and
(iv) to be adapted quickly to specific user needs ranging from single-skill
instruction to complete custom textbooks.
New authors, reviewers and field testers are welcome.
PROJECT STAFF
Andrew Schnepp
Eugene Kales
Peter Signell
Webmaster
Graphics
Project Director
ADVISORY COMMITTEE
D. Alan Bromley
E. Leonard Jossem
A. A. Strassenburg
Yale University
The Ohio State University
S. U. N. Y., Stony Brook
Views expressed in a module are those of the module author(s) and are
not necessarily those of other project participants.
c 2001, Peter Signell for Project PHYSNET, Physics-Astronomy Bldg.,
°
Mich. State Univ., E. Lansing, MI 48824; (517) 355-3784. For our liberal
use policies see:
http://www.physnet.org/home/modules/license.html.
3
4
MISN-0-210
1
ELECTROMAGNETIC WAVES FROM
MAXWELL’S EQUATIONS
MISN-0-210
2
another vector one.
by
Peter Signell
Michigan State University
1. Introduction
1a. Background. The transport of energy through mechanical systems via the collective motion of the particles that make up the system is
a familiar phenomenon, spectacularly demonstrated, for example, when
the voice of a soprano shatters a glass across the room from her. This
energy is carried by the “displacement” waves that can be made to propagate through the system (the “displacement” referring to the displacement
from equilibrium of the particles of the system or of the pressure in the
gas).
1b. Waves in Space. It is also possible for electric and magnetic
fields to propagate as waves in empty space, the electric and magnetic
field vectors playing the same role in electromagnetic waves as the transverse displacement of the particles of a string do in waves along a stretched
string, or the pressure displacement associated with the propagation of
sound waves in air. The important difference is that there is no medium
through which this electromagnetic wave propagates, although perhaps
one could say the “medium” is the vacuum! Electric and magnetic fields
may exist in space without a material medium being present, and if they
vary in space and time in the appropriate way, the spatial variation will
propagate as a wave, transporting energy. This module deals with the
propagation of energy through a vacuum via electromagnetic disturbances
whose space and time variation satisfy the conditions for wave propagation.
1c. Vector Derivatives. To deal with electromagnetic waves in space
it is far easier to use Maxwell’s equations in derivative form than in integral form. In electricity and magnetism we are dealing with scalar fields
~ and we will find that
like the charge density ρ and vector fields like E,
we must deal with three kinds of derivative operators: the “gradient”
operator that operates on a scalar field and produces a vector one, the
“divergence” operator that operates on a vector field and produces a scalar
one, and the “curl” operator that operates on a vector field and produces
5
1d. Partial Derivatives. When taking derivatives of field quantities
we generally use “partial derivatives,” denoted ∂, to remind us that the
spatial coordinates are not functions of time. That is, whereas a particle
has a single value of, say, x, at any particular time, a field has values at a
continuum of values of x. Otherwise, partial derivatives are like ordinary
derivatives:
∂ ¡ 3 4 5¢
x y z = 3x2 y 4 z 5 ,
∂x
∂
(sin 3x cos y) = 3 cos 3x cos y .
∂x
2. Vector Derivatives
2a. The Gradient Operator. The gradient operator operates on a
scalar function and produces a vector function that is the steepest “uphill” slope at any point where the vector function is evaluated. That is,
the gradient of a function, evaluated at some space-point, points in the
direction that is most steeply “up hill” in that function at that point. The
magnitude of the gradient is the value of the slope in the “steepest ascent”
direction at that space-point. As an example, suppose the gradient of a
field scalar field f is the vector field ~g . In Cartesian coordinates this is:
~ ≡ x̂ ∂f + ŷ ∂f + ẑ ∂f .
~g = ∇f
∂x
∂y
∂z
(1)
~ 3 y 4 ) = (3x2 y 4 )x̂ + (4x3 y 3 )ŷ.
¤ Show that ∇(x
2b. The Divergence Operator. The “divergence” operator operates
on a vector function, say ~g (x, y, z), to give a scalar function f (x, y, z):
~ · ~g ≡ ∂gx + ∂gy + ∂gz .
f =∇
∂x
∂y
∂z
(2)
The divergence of a vector function, evaluated at some space-point,
gives the extent to which the function has a source or sink at that point.
For example, the divergence of an electric field gives the charge density
at that point (positive charges are sources for the field, negative charges
are sinks).
¡
¢
~ · x3 y 4 z 5 ẑ = 5x3 y 4 z 4 .
¤ Show that: ∇
6
MISN-0-210
3
2c. The Curl Operator. The “curl” operator operates on a vector
function, say ~g (x, y, z), to give another vector function:
¶
µ
¶
µ
¶
µ
~ × ~g ≡ ∂gy − ∂gx ẑ + ∂gz − ∂gy x̂ + ∂gx − ∂gz ŷ . (3)
∇
∂x
∂y
∂y
∂z
∂z
∂x
The curl of a function, evaluated at some space-point, gives the greatest
“circulation” at that point, where by “circulation” one means the line
integral of the function around a loop of infinitesimal radius. The direction of the curl is normal to the plane of the loop with the greatest line
integral.1
¢
¡
¢
¡
¢
¡
~ × x3 y 4 z 5 ẑ = 4x3 y 3 z 5 x̂ − 3x2 y 4 z 5 ŷ.
¤ Show that: ∇
MISN-0-210
~
~
∂2B
~ × ∂E .
= −∇
2
∂t
∂t
We now put Eqs. (6) and (7) into the right sides of the above two equations
to get:
~
1 ∂2E
~ × (∇
~ × E)
~ ,
− 2 2 =∇
(8)
c ∂t
~
1 ∂2B
~ × (∇
~ × B)
~ ,
(9)
− 2 2 =∇
c ∂t
To further reduce the above equations, we make use of the identity:2
~ × (∇
~ × A)
~ = ∇(
~ ∇
~ · A)
~ − (∇
~ · ∇)
~ A
~ = ∇(
~ ∇
~ · A)
~ − ∇2 A
~.
∇
(10)
~ is any vector field and:
where A
3. Maxwell’s Equations
2
2
2
~ ·∇
~ = ∂ + ∂ + ∂ .
∇2 ≡ ∇
∂x2
∂y 2
∂z 2
Here are the famous Maxwell’s Equations in differential form:
~ ·E
~ = 4πke ρ ,
∇
(4)
~ ·B
~ = 0,
∇
~
~ ×E
~ = − ∂B ,
∇
∂t
(5)
Finally, using that identity and Eqs. (4) and (5) for our chargeless case,
we get these two wave equations for waves traveling with velocity c:
(6)
~
~ ×B
~ = 4πkm j + c−2 ∂ E .
(7)
∇
∂t
Gauss’s law is the integral form of Eq. (4). Ampere’s law is the integral
form of Eq. (7) for the case where the electric field does not vary with
time. The Ampere-Laplace-Biot-Savart law is derived from a combination
of Eqs. (5) and (7), also for the case where the electric field does not vary
with time. The Faraday-Henry law of magnetic induction is the integral
form of Eq. (6).
4. Electromagnetic Waves
4a. No Charge, No Current; Waves. For the case where there is
no charge or current at a point in space, it is easy to show that waves can
exist there. We set ρ = 0 and j = 0 in Eqs. (4)-(7), then take the time
derivative of both sides of Eqs. (6) and (7): Help: [S-3]
~
1 ∂ E
~ × ∂B ,
=∇
2
2
c ∂t
∂t
2~
1 For
4
~
1 ∂2B
~,
= ∇2 B
c2 ∂t2
(11)
~
1 ∂2E
~.
= ∇2 E
2
2
c ∂t
(12)
4b. Plane-Polarized Monochromatic Waves. We can write down
solutions to the wave equations, Eqs. (11) and (12), for the case where
the field vectors lie entirely in a plane and where the solutions contain
only one frequency. Of course one must show that the solutions we write
down really are solutions by substituting them into Eqs. (11) and (12) and
showing that those equations are satisfied. For the electric field vector the
plane-polarized monochromatic solution is:
~ =E
~ 0 cos(~k · ~r − ωt) ,
E
2 This
(13)
identity is proved in Appendix B. Physicists often remember the rule,
~ × (B
~ × C)
~ = B(
~ A
~ · C)
~ − C(
~ A
~ · B)
~ ,
A
as the words ”BAC minus CAB” along with the positions of the parentheses. However,
~ ≡ B
~ ≡ ∇,
~ we must rearrange the order on the right side of the equation to
with A
keep the operators to the left of the functions they operate upon. Thus we wind up
with Eq. (10).
another way to remember the definition of the curl, see Appendix A.
7
8
MISN-0-210
5
where the direction of ~k gives the direction of propagation of the electric
wave and the magnitude of ~k is 2π divided by the wave’s wavelength and
is related to the wave’s frequency through its velocity:3
k=
2π
;
λ
ω = 2πf ;
f=
1
c
= ,
T
λ
where ω is the wave’s angular frequency, f is its frequency, T is its period,
c is its speed, and λ is its wavelength.
¤ Show that Eq. (13) satisfies Eq. (12) by direct substitution on both sides
of Eq. (12). Help: [S-2]
Now with Eq. (13), Eq. (6) becomes:
~ ×E
~ = −(~k × E
~ 0 ) sin(~k · ~r − ωt) ,
∇
~ which can only be true if the arguments in the
but this equals (∂/∂t)B
cosine functions match:
~ =B
~ 0 cos(~k · ~r − ωt) .
B
~ 0 = 1 (~k × E
~ 0 ) = 1 (k̂ × E
~ 0) ,
B
ω
c
field to the right of the antenna is as given on this module’s cover. Then
use Eqs. (15) tell you the direction of the electric field and the direction
of propagation of the wave.
4d. How The Waves Manifest Themselves. Depending on its
frequency, an electromagnetic wave may be a radio or television wave
coming through the air to your receiver, or it could be an X-ray, or a
gamma ray from a radioactive decay, or a ray of light of a particular color.
These objects are all identical waves except for their frequencies.
Acknowledgments
Preparation of this module was supported in part by the National
Science Foundation, Division of Science Education Development and
Research, through Grant #SED 74-20088 to Michigan State University.
Glossary
• propagation: motion.
• polarization, direction of: in a plane-polarized electromagnetic
wave, the direction of the electric field vector.
so:
(15)
¤ Show that the picture on the cover of this module requires all three of
Eqs. (15).
4c. Production by a Radio Transmitter. A vertical radio transmitter tower is a good example of a device that produces a plane-polarized
monochromatic wave (the frequency of the wave is the frequency to which
you set the dial in order to receive the wave). A large current is sent up
and down the vertical tower, as a sine wave with a single frequency.
¤ Suppose we look at the tower during the part of the cycle when the
current is moving upward. Use Ampere’s law to show that the magnetic
3 These
6
(14)
We put the solutions, Eqs. (13) and (14), into Eqs. (11) and (12) and find:
Ê0 × B̂0 = k̂
~0 = 0
~0 · B
E
1
B 0 = E0 .
c
MISN-0-210
are general properties of waves: see “The Wave Equation,” (MISN-0-201).
9
• plane-polarization: in an electromagnetic wave, the condition in
which the electric field vector always lies in the same plane (in contrast to, say, circular polarization where the electric field vector rotates
around the axis of propagation).
• monochromatic: in an electromagnetic wave, the condition of a wave
having a single frequency, a single wavelength (in contrast to being a
mixture of different wavelengths). In a more sophisticated view, it
means that there is only one Fourier component.
A. The Curl as a Determinant
Recall that the cross-product
as the determinant
¯
¯
¯
~ ×D
~ =¯
C
¯
¯
~ and D
~ can be written
of two vectors C
x̂
Cx
Dx
ŷ
Cy
Dy
ẑ
Cz
Dz
¯
¯
¯
¯
¯
¯
10
MISN-0-210
7
MISN-0-210
~ ×D
~ is the term in the expanded determinant
The x-component of vector C
which is proportional to x̂:
~ × D)
~ x = ∂Dz − ∂Dy
(∇
∂y
∂z
PS-1
PROBLEM SUPPLEMENT
¤ Warning: First make sure you have done the first five (out of the six)
problems scattered through the text, marked like this warning. If you
skip any one of them, you will probably not be prepared for even the first
problem below.
The other two components of the curl of D are thus:
~ × D)
~ y = ∂Dx − ∂Dz
(∇
∂z
∂x
Note: Problems 4-7 also occur in this module’s Model Exam.
~ × D)
~ z = ∂Dy − ∂Dx
(∇
∂x
∂y
1. The electric field of a plane electromagnetic wave in vacuum is represented by:
Ex = 0 ,
B. Proof of a Vector Identity
£
¤
Ey = 0.50 (N/C) cos 2.09 m−1 (x − ct) ,
We here prove:
Ez = 0 .
~ × (B
~ × C)
~ = −(A
~ · B)
~ C
~ + B(
~ A
~ · C)
~ .
A
We only need to prove the identity for one component since the others
will follow by cycling the subscripts. So we take the x-component of the
left side:
~ × (B
~ × C)]
~ x
[A
a. Determine the wavelength, frequency, polarization, and propagation
vector of the wave. Help: [S-4]
b. Determine the
Help: [S-1]
components
of
the
wave’s
magnetic
field.
2. Solve (a) and (b) of Problem 1 for the wave represented by:
~ × C)
~ z − A z (B
~ × C)
~ y
= A y (B
Ex = 0 ,
= Ay (Bx Cy − By Cx ) − Az (Bz Cx − Bx Cz )
£
¤
Ey = 0.50 (N/C) cos 0.419 m−1 (x − ct) ,
£
¤
Ez = 0.50 (N/C) cos 0.419 m−1 (x − ct) .
= −(Ay By + Az Bz )Cx + Bx (Ay Cy + Az Cz )
= −(Ax Bx + Ay By + Az Bz )Cx + Bx (Ay Cy + Az Cz + Ax Cx )
~ · B)C
~ x + B x (A
~ · C)
~
= −(A
~ and B-fields
~
3. Determine the components of the Ewhich describe the
following electromagnetic waves that propagate along the positive xaxis:
hence:
~ × (B
~ × C)
~ = −(A
~ · B)
~ C
~ + B(
~ A
~ · C)
~
A
~
a. A wave whose plane of E-vibration
makes an angles of 45◦ with the
positive y- and z-axes.
~
b. A wave whose plane of E-vibration
makes an angle of 120◦ with the
positive y axis and an angle of 30.0◦ with the positive z-axis.
~ for A
~ and B:
~
so substituting ∇
~ × (∇
~ × C)
~ = −(∇
~ · ∇)
~ C
~ + ∇(
~ ∇
~ · C)
~
∇
~ × (∇
~ × C)
~ = −∇2 C
~ + ∇(
~ ∇
~ · C)
~ ,
∇
and the identity is proved.
11
12
MISN-0-210
PS-2
4. Given these electric and magnetic fields:
Ex = E0 cos
Bx = 0,
2π
(y + ct),
λ
Ey = 0,
Ez = 0,
By = 0,
Bz = 0.
MISN-0-210
PS-3
Brief Answers:
1. a. 3.01 m, 1.00 × 108 Hz, polarized in the ŷ direction, ~k = 2.09 m−1 x̂.
b. Bx = 0 ,
a. Determine whether or not these fields satisfy the wave equations.
[D]
b. Determine whether or not these fields satisfy Maxwell’s Equations.
[B]
c. If your answer to (a) and (b) is yes, what relationship must exist
between the E and B amplitudes? [A]
By = 0 ,
£
¤
Bz = 0.17 × 10−8 T cos 2.09 m−1 (x − ct) .
2 a. 15.0 m, 2.00 × 107 Hz, polarized in the y-z plane, ~k = 0.419 m−1 x̂ .
b. Bx = 0 ,
By = −0.17 × 10−8 T cos[0.419 m−1 (x − ct)] ,
Bz = +0.17 × 10−8 T cos[0.419 m−1 (x − ct)] ,
5. Given these electric and magnetic fields:
³x
´
Ex = 0, Ey = E0 sin[2πν
− t ], Ez = 0,
c
´
³x
Bx = 0, By = 0,
− t ].
Bz = B0 sin[2πν
c
3 a. Ex = 0,
Ey = +0.707E0 cos(kx − ωt).
Ez = +0.707E0 cos(kx − ωt).
Bx = 0,
(a), (b), (c): Repeat Problem 1 using the above components.
By = −0.707(E0 /c) cos(kx − ωt).
Answers: (a) [G], (b) [E], and (c) [H].
Bz = +0.707(E0 /c) cos(kx − ωt).
b. Ex = 0,
6. With:
Ey = −0.500E0 cos(kx − ωt),
Ex = E0 cos(kz − ωt), Ey = E0 cos(kz − ωt), Ez = 0,
write down the space-time dependence of the components of the magnetic field that will result in an electromagnetic wave that satisfies
both the wave equations and Maxwell’s equations. [C]
7. A plane-polarized monochromatic electromagnetic wave of frequency
ν has the electric field polarized in the z-direction and the wave propa~ and
gates in the negative y-direction. Determine the components of E
~
B that satisfy Maxwell’s equations and the wave equations. [F]
Ez = +0.866E0 cos(kx − ωt).
Bx = 0,
By = −0.866(E0 /c) cos(kx − ωt).
Bz = −0.500(E0 /c) cos(kx − ωt),
A. No such wave exists.
B. No.
C. Bx = −
E0
cos(kz − ωt),
c
By = +
E0
cos(kz − ωt),
c
Bz = 0.
D. Yes.
E. Yes.
13
14
MISN-0-210
PS-4
MISN-0-210
AS-1
F. Ex = 0,
SPECIAL ASSISTANCE SUPPLEMENT
Ey = 0,
h
Ez = +E0 cos 2πν
³y
´i
+t ,
S-1
c
³
´i
E0
y
cos 2πν
+t ,
Bx = −
c
c
(from PS-Problem 1)
For electromagnetic waves, the three important directions are: k̂, the
direction of propagation; Ê, the direction of the electric field; and B̂,
the direction of the magnetic field. Any two of these may be known in
a problem and we must find the third. Since these three are mutually
perpendicular, they obey this cyclic rule:
h
By = 0,
Bz = 0.
G. Yes.
Ê × B̂ = k̂ ,
H. B0 = E0 /c.
k̂ × Ê = B̂ ,
B̂ × k̂ = Ê ,
where each line is obtained from the one above it by cycling the vectors
one place to the right.
S-2
(from TX-4b)
Recall that ~r = xx̂ + y ŷ + z ẑ and ~k · ~r = kx x + ky y + kz z. Then:
~ x (~k · ~r) = ∂ (~k · ~r) = kx .
∇
∂x
~ y and ∇
~ z . Then:
and similarly for ∇
~ ~k · ~r) = ~k .
∇(
This means that:
~ cos(~k · ~r − ωt) = −~k sin(~k · ~r − ωt) .
∇
Now you fill in the remaining steps to get:
~ = −k 2 E
~.
∇2 E
and do a similar job on the other side of the wave equation.
15
16
MISN-0-210
S-3
AS-2
MISN-0-210
ME-1
(from TX-4a)
The various partial derivatives, (∂/∂t), (∂/∂x), etc., are independent of
each other so can be taken in any order. Thus, for example,
MODEL EXAM
~ ∂ f (~r, t) = ∂ ∇f
~ (~r, t) ,
∇
∂t
∂t
~ = x̂ ∂f + ŷ ∂g + ẑ ∂h
∇f
∂x
∂y
∂z
~ · ~g = ∂gx + ∂gy + ∂gz
∇
∂x
∂y
∂z
where f is any function.
S-4
(from PS-Problem 1)
(~k·~r) is written in the problem as (2.09q
m−1 x). The obvious conclusion is
that ky = kz = 0 and hence that k = kx2 + ky2 + kz2 = kx = 2.09 m−1 .
~ × ~g =
∇
µ
∂gy
∂gx
−
∂x
∂y
¶
ẑ +
µ
∂gy
∂gz
−
∂y
∂z
¶
x̂ +
µ
∂gz
∂gx
−
∂z
∂x
¶
ŷ
~ ∇
~ · ~g ) ; ∇2 ≡ ∇
~ ·∇
~
~ × (∇
~ × ~g ) = −∇2~g + ∇(
∇
~ ·E
~ = 4πke ρ
∇
~ ·B
~ =0
∇
~
~ ×E
~ = − ∂B
∇
∂t
~
~ ×B
~ = 4πkm j + c−2 ∂ E
∇
∂t
1. See Output Skills K1-K2 in this module’s ID Sheet.
2. Given these electric and magnetic fields:
Ex = E0 cos
Bx = 0,
2π
(y + ct),
λ
Ey = 0,
Ez = 0,
By = 0,
Bz = 0.
a. Determine whether or not these fields satisfy the wave equations.
b. Determine whether or not these fields satisfy Maxwell’s Equations.
c. If your answer to (a) and (b) is yes, what relationship must exist
between the E and B amplitudes?
3. Given these electric and magnetic fields:
´
³x
− t ], Ez = 0,
Ex = 0, Ey = E0 sin[2πν
c
´
³x
− t ].
Bz = B0 sin[2πν
Bx = 0, By = 0,
c
(a), (b), (c): Repeat Problem 1 using the above components.
17
18
MISN-0-210
ME-2
4. With:
Ex = E0 cos(kz − ωt), Ey = E0 cos(kz − ωt), Ez = 0,
write down the space-time dependence of the components of the magnetic field that will result in an electromagnetic wave that satisfies
both the wave equations and Maxwell’s equations.
5. A plane-polarized monochromatic electromagnetic wave of frequency
ν has the electric field polarized in the z-direction and the wave propa~ and
gates in the negative y-direction. Determine the components of E
~ that satisfy Maxwell’s equations and the wave equations.
B
Brief Answers:
1. See this module’s text.
2. See Problem 4 in this module’s Problem Supplement.
3. See Problem 5 in this module’s Problem Supplement.
4. See Problem 6 in this module’s Problem Supplement.
5. See Problem 7 in this module’s Problem Supplement.
19
20